Evaluate $\int\tan x\,\sec^3x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac13\tan x\sec^3x+C$ (Choice B) B $\dfrac14\tan x\sec^4x+C$ (Choice C) C $\dfrac13\sec^3x+C$ (Choice D) D $\dfrac14\sec^4x+C$
Explanation: In this problem we have an odd power of $\tan x$ and some powers of $\sec x$. We note that the derivative of $\sec x$ is $\sec x\,\tan x$. Hence we rewrite the integral as $\int \tan{x}\, \sec^{3}x \ dx = \int (\sec x\,\tan x)\ \sec^2x \ dx\,$ Now we use a $u$ -substitution with $u=\sec x$ and $du =\tan x\sec x\,dx$. $ \int (\sec x\,\tan x)\ \sec^2x \ dx= \int u^{2} \ du$ Integrating and plugging $\sec x$ back in for $u$, we get $ \int u^{2} \ du=\dfrac{u^3}3+C=\dfrac13\sec^3x+C\,$